Energy Work Problem, all algebra

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SUMMARY

The discussion focuses on solving a physics problem involving energy conservation and projectile motion. The block is released from height h on a frictionless ramp inclined at angle theta, and the goal is to determine the height h required for the block to land in a hole located a horizontal distance x from the ramp's end and a vertical distance y below it. The key equation derived is mg(h+y)=(1/2)mv^2, leading to h=(v^2/2g)-y. The challenge lies in finding the expression for v^2, as it involves eliminating h from both sides of the equation.

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enkerecz
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Homework Statement



A small block is placed at height h on a frictionless ramp inclined at an angle theta. Upon being released, the block slides down the ramp and then falls to the floor (a distance y below the base of the ramp). A small hole is located a horizontal distance x from the end of the ramp.

From what height, h , should the block be released in order to land in the hole? Note: that the unknowns for this problem are x, y, and theta. Your answer should be an algebraic expression that starts with h=


Homework Equations





The Attempt at a Solution



I started out by using the fact that
mg(h+y)=(1/2)mv^2
working it out to
h= (v^2/2g)-Y

my issue is finding v^2, I know that (2gh)^1/2 is the final velocity for both mgh and mgy.. I need to eliminate the h somehow because it is on both sides, but I am unsure how to do so. I have used trig, but am having no luck. I already solved this problem once using kinematics, and found it to be much easier. Any help solving this with energy would be appreciated.
 
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Welcome to PF!

Hi enkerecz! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)
enkerecz said:
… I already solved this problem once using kinematics, and found it to be much easier. Any help solving this with energy would be appreciated.

(how did you manage to solve this without using x ? :confused:)

I don't think energy will give you anything more than the velocity on leaving the ramp: after that, you will have to use the standard constant acceleration equations. :smile:
 

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